CODEWITHSUNDEEP
rdate
Nigel
Nigel

Reputation: 3

How to convert a date with only a year to a date with the format "Year-Month-Day" in R

Sorry for the question, I started using RStudio a month ago and I get confronted to things I've never learned. I checked all the websites, helps and forums possible the past two days and this is getting me crazy.

I got a variable called Release giving the date of the release of a song. Some dates are following the format %Y-%m-%d whereas some others only give me a Year. I'd like them to be all the same but I'm struggling to only modify the observations with the year.

Brief summary in word:

11/11/2011
01/06/2011
1974
1970
16/09/2003

I've imported the data with :

music<-read.csv("music2.csv", header=TRUE, sep = ",", encoding = "UTF-8",stringsAsFactors = F)

And this how I have it in RStudio

"2011-11-11" "2011-06-01" "1974" "1970" "2003-09-16"

This is an example as I got 2200 obs.

The working code is 

Modifdates<- ifelse(nchar(music$Release)==4,paste0("01-01-",music$Release),music$Release)
Modifdates

I obtain this : 

"2011-11-11" "2011-06-01" "01-01-1974" "01-01-1970" "2003-09-16"

I just would like them to be all with the same format "%Y-%m-%d". How can I do that?

So I tried this

as.Date(music$Release,format="%Y-%m-%d")

But I got NA's where I modified my dates.

Could anyone help?

Upvotes: 0

Views: 332

Answers (2)

utubun
utubun

Reputation: 4520

Update

Using sub find occurrences of date consisting from single year ("(^[0-9]{4}$)" part), using back-reference substitute it to add -01-01 at the end of the string ("\\1-01-01" part), and finally convert it to the date class, using as.Date() (as.Date() default is format = "%Y-%m-%d" so you don't need to specify it):

dat <- c("2011-11-11", "2011-06-01", "1974", "1970", "2003-09-16")
dat class is character: 
as.Date(sub("(^[0-9]{4}$)", "\\1-01-01", dat))

# "2011-11-11" "2011-06-01" "1974-01-01" "1970-01-01" "2003-09-16"
dat class is factor, but sub automatically coerce it to the character class for you: 
# dat <- as.factor(dat); dat

# 2011-11-11 2011-06-01 1974       1970       2003-09-16
# Levels: 1970 1974 2003-09-16 2011-06-01 2011-11-11

as.Date(sub("(^[0-9]{4}$)", "\\1-01-01", dat))

# "2011-11-11" "2011-06-01" "1974-01-01" "1970-01-01" "2003-09-16"

Upvotes: 1

gaut
gaut

Reputation: 5958

Welcome to SO, please try to provide a reproducible example next time so that we can best help you. I think here you could use:

testdates <- c("1974", "12-12-2012")
betterdates <- ifelse(nchar(testdates)==4,paste0("01-01-",testdates),testdates)
> betterdates
[1] "01-01-1974" "12-12-2012"

EDIT: if your vector is factor you should use as.character.factor first. If you then want to convert back to factor you can use as.factor

EDIT2 : do not convert as.date before doing this. Only do it after this modification

Upvotes: 1

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